Exercise 2.11-Additional Questions - Chapter 2 Basic Algebra 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions Solved
Question 1.
Simplify $(343)^{2 / 3}$
Solution:
$
\begin{aligned}
& 343=7^3 \\
& \text { So }(343)^{2 / 3}=\left(7^3\right)^{2 / 3}=7^{3 \times \frac{2}{3}}=7^2=49
\end{aligned}
$
Question 2.
Simplify: $\frac{1}{2+\sqrt{3}}+\frac{3}{4-\sqrt{5}}+\frac{6}{7-\sqrt{8}}$

Solution:
$
\begin{gathered}
\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3} \\
\frac{3}{4-\sqrt{5}}=\frac{3}{4-\sqrt{5}} \times \frac{4+\sqrt{5}}{4+\sqrt{5}}=\frac{12+3 \sqrt{5}}{16-5}=\frac{12+3 \sqrt{5}}{11} \\
\frac{6}{7-\sqrt{8}}=\frac{6}{7-\sqrt{8}} \times \frac{7+\sqrt{8}}{7+\sqrt{8}}=\frac{42+6 \sqrt{8}}{49-8}=\frac{42+6 \sqrt{8}}{41} \\
\therefore \frac{1}{2-\sqrt{3}}+\frac{3}{4-\sqrt{5}}+\frac{6}{7-\sqrt{8}}=2-\sqrt{3}+\frac{12+3 \sqrt{5}}{11}+\frac{42+6 \sqrt{8}}{41}
\end{gathered}
$
Question 3.
Simplify: $\sqrt{98}+\sqrt{50}-\sqrt{18}+\sqrt{75}-\sqrt{27}$
Solution:
$
\begin{aligned}
& \sqrt{98}=\sqrt{49 \times 2}=7 \sqrt{2} ; \sqrt{18}=\sqrt{9 \times 2}=3 \sqrt{2} \\
& \sqrt{50}=\sqrt{25 \times 2}=5 \sqrt{2} ; \sqrt{75}=\sqrt{25 \times 3}=5 \sqrt{3} \\
& \sqrt{27}=\sqrt{9 \times 3}=3 \sqrt{3} \\
& \therefore \quad \sqrt{98}+\sqrt{50}-\sqrt{18}+\sqrt{75}-\sqrt{27}=7 \sqrt{2}+5 \sqrt{2}-3 \sqrt{2}+5 \sqrt{3}-3 \sqrt{3} \\
&=9 \sqrt{2}+2 \sqrt{3}
\end{aligned}
$
Question 4.
Solve $(1-x)^{1 / 4}+(15+x)^{1 / 4}=2$
Solution:

$
(1-x)^{\frac{1}{4}}+(15+x)^{\frac{1}{4}}=2
$
squaring on both sides
$
\begin{gathered}
(1-x)^{\frac{1}{2}}+(15+x)^{\frac{1}{2}}+2[(1-x)(15+x)]^{\frac{1}{4}}=4 \\
(1-x)^{\frac{1}{2}}+(15+x)^{\frac{1}{2}}=4-2[(1-x)(15+x)]^{\frac{1}{4}}
\end{gathered}
$
squaring on both sides again
$
\begin{aligned}
& (1-x)+(15+x)+2 \sqrt{(1-x)(15+x)}=16+4(1-x)^{\frac{1}{2}}(15+x)^{\frac{1}{2}}-16[(1-x)(15+x)]^{\frac{1}{4}} \\
& 1-x+15+x+2 \sqrt{(1-x)(15+x)}=16+4 \sqrt{(1-x)(15+x)}-16[(1-x)(15+x)]^{\frac{1}{4}} \\
& 16+2 \sqrt{(1-x)(15+x)}-16-4 \sqrt{(1-x)(15+x)}=-16[(1-x)(15+x)]^{\frac{1}{4}} \\
& -2 \sqrt{(1-x)(15+x)}=-16[(1-x)(15+x)]^{\frac{1}{4}} \\
& \sqrt{(1-x)(15+x)}=8[(1-x)(15+x)]^{\frac{1}{4}}
\end{aligned}
$
squaring on bothsides

$
\begin{aligned}
& (1-x)+(15+x)+2 \sqrt{(1-x)(15+x)}=16+4(1-x)^{\frac{1}{2}}(15+x)^{\frac{1}{2}}-16[(1-x)(15+x)]^{\frac{1}{4}} \\
& 1-x+15+x+2 \sqrt{(1-x)(15+x)}=16+4 \sqrt{(1-x)(15+x)}-16[(1-x)(15+x)]^{\frac{1}{4}} \\
& 16+2 \sqrt{(1-x)(15+x)}-16-4 \sqrt{(1-x)(15+x)}=-16[(1-x)(15+x)]^{\frac{1}{4}} \\
& -2 \sqrt{(1-x)(15+x)}=-16\left[(1-x)(15+x)^{\frac{1}{4}}\right. \\
& \sqrt{(1-x)(15+x)}=8[(1-x)(15+x)]^{\frac{1}{4}}
\end{aligned}
$
squaring on both sides again
$
(1-x)(15+x)=64[(1-x)(15+x)]^{\frac{1}{2}}
$
squaring on both sides again
$
\begin{aligned}
& {[(1-x)(15+x)]^2=64(1-x)(15+x) } \\
& \div \text { by }(1-x)(15+x) \text { we get } \\
&(1-x)(15+x)=64 \\
& 15-15 x-x^2+x-64=0 \\
&-x^2-14 x-49=0 \\
& \Rightarrow \quad x^2+14 x+49=0 \\
&(x+7)^2=0 \Rightarrow x=-7
\end{aligned}
$
Question 5.
Solve $(x+1)^{\frac{1}{3}}=\sqrt{x-3}$
Solution:
$
(x+1)^{\frac{1}{3}}=(x-3)^{\frac{1}{2}}
$
L.C.M. of 2 and 3 is $6 \&$ Raising to the power 6

$
\begin{aligned}
& \left\{(x+1)^{\frac{1}{3}}\right\}^6=\left\{(x-3)^{\frac{1}{2}}\right\}^6 \\
& (x+1)^2=(x-3)^3 \\
& x^2+2 x+1=x^3-9 x^2+27 x-27 \\
& 0=x^3-9 x^2+27 x-27-x^2-2 x-1 \\
& x^3-10 x^2+25 x-28=0
\end{aligned}
$
since constant term is -28
we can have a factor as $(x \pm 2)$ or $(x \pm 4)$ or $(x \pm 7)$
By trial and error method we find that $(x-7)$ is a factor
Using synthetic division

we get $x^3-10 x^2+25 x-28=(x-7)\left(x^2-3 x+4\right)$
solving $x^2-3 x+4=0$
$
x=\frac{3 \pm \sqrt{9-16}}{2}=\frac{3 \pm \sqrt{-7}}{2}
$
The roots are $x=7, x=\frac{3 \pm \sqrt{-7}}{2}$